If $X$ and $Y$ are Riemann surfaces (not necessarily compact), and $f:X\to Y$ is a holomorphic function, then it is obvious that the ramification points of $f$ in $X$ form a discrete subset of $X$. Is the same true of the branch points of $f$ (the set made up of the images of the ramification points)?
Dear Robert, there exists a holomorphic function $X\to Y $ having non discrete and even dense set of branch points, with $X=\mathbb C^\ast \setminus \{0\}$ and $Y=\mathbb C$. Bibliography and comments The version of MittagLeffler used above is not so easy to find in the literature (I just checked). It is proved in AshNovinger's Complex Variables ( theorem 6.3.3 ) where they deduce from it some algebraic properties ( due to Helmer) of the ring $\mathcal O(D)$ of holomorphic functions on an open connected $D\subset \mathbb C$. It is a nonnoetherian domain, not a UFD but any collection of elements of $\mathcal O(D)$ has a GCD and all its finitely generated ideals are principal. 


This is true if $f$ is proper (the preimage of a compact is compact). Indeed, any compact neighborhood of $y\in Y$ contains only finitely many branch points because its preimage in $X$ contains only finitely many ramification points. If $f$ is not proper, ramification points in $Y$ may not be discrete. Take for $X$ the union of copies of $\mathbb C$ indexed by $n=1,2,3,...$ and let $f:X\to \mathbb C$ be the holomorphic map which on the $n$th copy of $\mathbb C$ in $X$ is given by $f(z) = (z1/n)^2$. 


No: $f: \mathbb{C}\setminus \{0\} \to \mathbb{C}$, $f(z):= \frac{1}{z} sin (z)$. 


Here is another example using only algebraic and not transcendental functions. Take three distinct complex numbers $z_1,z_2,z_3$ and let $X$ be the Riemann surface obtained by integrating the form $\sqrt{(zz_1)(zz_2)(zz_3)}dz$. The map $g: X \to\mathbb C$ which sends each germ $\varphi_z\in X$ to $z$ is a covering with infinitely many sheets above the $z$plane. (The ramification points are exactly above $z_1,z_2,z_3$, and there are infinitely many; each one is of order 2. This is so because near $z_k$ the form is like $(zz_k)^{1/2}dz$ and therefore the primitive is like $(zz_k)^{3/2}+cte$.) Now consider the holomorphic map $f: X\to\mathbb C$ which sends each germ $\varphi_z\in X$ to $\varphi_z(z)$. The branch points of $f$ are exactly the same as the branch points of $g$, but whilst they are of order 2 for $g$ they are of order 3 for $f$. (To see why this is so resolve the equation $\zeta=(zz_k)^{3/2}+cte$ in $z$.) Now comes the important point: For nearly all choices of $z_1,z_2,z_3$ the periods $a_1,a_2,a_3$ of $ydz$ on the elliptic curve $y^2=(zz_1)(zz_2)(zz_3)$ are such that the lattice $\mathbb{Z}a_1+\mathbb{Z}a_2+\mathbb{Z}a_3$ is dense in the plane. This means that the set of the images of the branch points of $f$ is dense in the plane. In this way you can construct many other examples... 


Another very concrete answer but without any formula: I suppose you know how to construct "with papers, scissors and glue" a Riemann surface with a single branch point of order 2 (for example the surface of the function $\sqrt z$). Now take the Riemann surface of the logarithm. It has a countable infinite number of sheets. On each sheet you can add "with papers, scissors and glue" a branch point of order two, and this at any place you wish except above the origin. In this way you construct, for any given countable set $A\subset\mathbb C$ a Riemann surface $f : X \to \mathbb C$ which has a branch point above every point of A. Moreover you see in the same way, that you can prescribe any (finite or infinite) order to each branch point (just glue more sheets, as you would do for $\sqrt[n]z$ or $\ln$); and you can also prescribe the number of branch points you want to have above each point of A (above each point of A you may want to have a countable number of distinct branch points). 

